Question

Please help differentiation

Answer 1

\[\sqrt{1+sinx \div1-sinx}\]

Answer 2

You can also think of it like this. This might make it easier. You will be using chain rule with in quotient rule.
\[\large\left[ \frac{ 1+\sin(x) }{ 1-\sin(x) } \right]^\frac{ 1 }{ 2 } \]

\(\huge OR\)
\[\large \left[ (1+\sin(x)) \times (1-\sin(x))^{-1} \right]^\frac{ 1 }{ 2 }\]

and use chain rule and product rule. Its all whichever you feel comfortable using.

Answer 3

i did that but am getting only secx but the answer is secx(secx+tanx)
please do the steps please

Answer 4

@xavier123 and @waterineyes help please

Answer 5

\[y=\sqrt{\frac{1+sinx}{1-sinx}}\]?

Answer 6

Power+Chain+Quotient rule

Answer 7

\[y'=(\frac{1+sinx}{1-sinx})^{\frac{1}{2}} \\ \\ y'=\ \frac{1}{2}(\frac{1+sinx}{1-sinx})^{-\frac{1}{2}}(\frac{d}{dx}(\frac{1+\sin x}{1-\sin x}))\]

Answer 8

\[y'=\ \frac{1}{2}(\frac{1+sinx}{1-sinx})^{-\frac{1}{2}}(\frac{2 ( \cos x )}{( \sin x-1)^2})\]
Agree?

Answer 9

So...
\[\Large y'=\frac{1}{\sqrt{\frac{1+ \sin x }{1- \sin x }}}(\frac{\cos x}{(sinx -1)^2})\]

Answer 10

Agree?

Answer 11

Hey, @.Sam. just a query... and by the way, @yashdhanuka if this confuses you, you don't have to look at this, but by all means :)

\[\large \sqrt{\frac{1+\sin(x)}{1-\sin(x)}}=\left(\sqrt{\frac{1-\cos(\frac{\pi}2-x)}{1+\cos(\frac\pi 2 - x)}}\right)^{-1}=\left[\tan\left(\frac \pi 4 - \frac x 2\right)\right]^{-1}\]
\[\Large = \cot\left(\frac \pi 4 - \frac x 2\right)\]

Answer 12

answer is secx(secx+tanx) :(

Answer 13

multiply and divide by sqrt(1+sin(x))...
u get (1+sin(x))/|cos(x)|... take into two parts... when cos(x) > 0 and when cos(x) <0...